Problem: Simplify the following expression and state the condition under which the simplification is valid. $z = \dfrac{a^2 - 64}{a + 8}$
Solution: First factor the polynomial in the numerator. The numerator is in the form ${a^2} - {b^2}$ , which is a difference of two squares so we can factor it as $({a} + {b})({a} - {b})$ $ a = a$ $ b = \sqrt{64} = 8$ So we can rewrite the expression as: $z = \dfrac{({a} + {8})({a} {-8})} {a + 8} $ We can divide the numerator and denominator by $(a + 8)$ on condition that $a \neq -8$ Therefore $z = a - 8; a \neq -8$